Exercise 2.4.1 (Cofinite topology is separable, not necessarily second countable)

Proof. If $X$ is finite, then $X$ itself is a countable subset with a closure equal to $X$. Thus, assume that $X$ is infinite. Then the only closed sets in $X$ (except for $X$ itself) are those with finite cardinality. Closure of any countable subset of $X$ is thus $X$ itself. □

Proof. Since $X$ is countable, $X$ has only a countable number of finite spaces¹ . Since there is a one-to-one relationship between the open sets and the finite sets, we conclude that the set of all open sets forms a countable basis for the cofinite topology of $X$. □

Proof. Suppose for the sake of contradiction that there is a countable basis $B$ for the cofinite topology of $X$. Consider the set $B:={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{U\in B}X\setminus U$. Since the sets $X\setminus U$, $U\in B$, are finite, $B$ must be at most countable. Thus, we can pick a $x\in X\setminus B$. The set $X\setminus \{x\}$ is obviously open; there is, however, no $U\in B$ completely contained in $X\setminus \{x\}$ - a contradiction to Theorem 4.1. □